# Misc 15 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4π/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let β BOC = ΞΈ Now, AC = AO + OC H = r + r cos ΞΈ And, R = r sin ΞΈ H = r (1 + cos ΞΈ) We need to maximize volume of cone. Volume of the cone is V = 1/3 ππ ^2 π» V = 1/3 ππ^2 sin^2β‘"ΞΈ" r (1 + cos ΞΈ) V = 1/3 ππ^3 sin^2β‘"ΞΈ" (1 + cos ΞΈ) Finding ππ£/ππ ππ£/ππ = 1/3 ππ^3 [2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )+sin^2β‘γ"ΞΈ" (βsinβ‘γ"ΞΈ" )γ γ γ γ γ ] ππ£/ππ = 1/3 ππ^3 (2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )βsin^3β‘"ΞΈ" γ γ γ ) ππ£/ππ = 1/3 ππ^3 (2 sinβ‘γ"ΞΈ" cosβ‘γ"ΞΈ" (1+cosβ‘γ"ΞΈ" )βsin^3β‘"ΞΈ" γ γ γ ) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (2 cosβ‘γ"ΞΈ" +γ2 cos^2γβ‘γ"ΞΈ" βsin^2β‘"ΞΈ" γ γ ) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (2 cosβ‘γ"ΞΈ" +γ2 cos^2γβ‘γ"ΞΈ" β(1βcos^2β‘"ΞΈ" γ γ)) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 cos^2β‘γ"ΞΈ" +γ2 πππ γβ‘γ"ΞΈ" β1γ γ ) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 cos^2β‘γ"ΞΈ" +γ3 πππ γβ‘π γ βcosβ‘π β1) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 γπππ γβ‘"ΞΈ" (cosβ‘π+1)β1(cosβ‘π+1)) ππ£/ππ = 1/3 ππ^3 sin ΞΈ (3 πππ β‘γ"ΞΈ" β1γ )(πππ β‘γ"ΞΈ" +1γ ) Putting ππ£/ππ = 0 1/3 ππ^3 sin ΞΈ (πππ β‘γ"ΞΈ" +1γ ) (3 πππ β‘γ"ΞΈ" β1γ ) = 0 sin ΞΈ (cos ΞΈ + 1) (3 cos "ΞΈ" β 1) = 0 π¬π’π§ ΞΈ = 0 ΞΈ = 0Β° ΞΈ cannot be 0Β° for cone ππ¨π¬ ΞΈ + 1 = 0 cos ΞΈ = β1 For cone, 0Β° < ΞΈ < 90Β° & cos ΞΈ is negative in II & III quadrant. So cos ΞΈ = β1 is not possible π ππ¨π¬ ΞΈ β 1 = 0 cos ΞΈ = 1/3 ΞΈ = cosβ1 1/3 cos ΞΈ = 1/3 is possible So, πππ β‘"ΞΈ" = 1/3 Thus, H = r (1 + cos ΞΈ) H = r ("1 + " 1/3) H = 4π/3 This value is either maxima or minima. So, finding (π^2 π)/(ππ^2 ) ππ/ππ = (1/3 ππ^3 sinβ‘π (3 cos2 π + 2 cosβ‘πβ 1)) = 1/3 ππ^3 [cosβ‘π (3 cos^2β‘π+2 cosβ‘πβ1)+sinβ‘π (6 cosβ‘π (βπ ππ)β2 sinβ‘π )] = 1/3 ππ^3 [3 cos^3β‘π+2 cos^2β‘πβcosβ‘π+sinβ‘π (β6 cosβ‘π sinβ‘πβ2 sinβ‘π )] = 1/3 ππ^3 [3 cos^2β‘π+2 cos^2β‘πβcosβ‘πβ6 sin^2β‘π cosβ‘πβ2 sin^2β‘π ] Now, cosβ‘π=1/3 And sin2 "ΞΈ" = 1 β cos2 "ΞΈ" = 1 β (1/3)^2= 1 β 1/9 = 8/9 Putting values in (π^2 π)/(ππ^2 ) (π^2 π)/(ππ^2 ) = 1/3 ππ^3 [3(1/3)^3+2(1/3)^2β1/3β6(1/3)(8/9)β2(8/9)] = 1/3 ππ^3 [3(1/27)+2(1/9)β1/3β6(8/27)β2(8/9)] = 1/3 ππ^3 [1/9+ 2/9β1/3β16/9β16/9] = 1/3 ππ^3 [(β32)/9] = (β32ππ^3)/27 Thus, (π^2 π£)/(ππ^2 ) < 0 H = ππ/π Hence proved

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.